In the square ABCD, the angle pad = angle PDA = 15 degrees, proving that the triangle PBC is an equilateral triangle

In the square ABCD, the angle pad = angle PDA = 15 degrees, proving that the triangle PBC is an equilateral triangle

Make the vertical line of AD through P, intersect ad to e, intersect BC to F,
Because PA = PD, so AE = ed, so EF is the vertical bisector of AD, EF is also the vertical bisector of BC, so Pb = PC, triangle PBC is isosceles triangle,
Let the side length of the square be a, PE = AE * tan15 = A / 2 * tan15
So pf = A-A / 2 * tan15,
tanPBC=PF/BF=[a-a/2*tan15]/(a/2)=[a-a/2*(2-√3）]/(a/2)=√3
So the angle PBC = 60 degrees, so the triangle PBC is an equilateral triangle